New lower price for Axon II ($78) and Axon Mote ($58).
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There were a lot of different kinds to choose from. Attached is a picture of the pack I ended up getting. It cost somewhere around $1. The top is a bit torn from opening it, but it said "These transistors are designed for high-speed, medium-power switching and general purpose amplifier applications."
So I get home and get out my multimeter and set it to the 200ohms range. I connect the + side of the multimeter to the base and the other side to one of the side pins. The meter jumps to 71ohms then goes to open circuit. I move the negative side of the meter to the other pin. Still nothing. I decide to switch and put the negative side of the meter on the base and try the positive side on the other pins. Nothing happens at all. I try changing the range of the meter; still nothing. I just can't get the transistors to do anything. I tried a few different transistors, all the with the same results.
with a gain of 120, in order to switch it on I need to use 800/120 = ~7mA?
How much voltage and current can I send through the collector-emitter portion of the transistor without killing it?
What is the minimum and maximum voltage and current I need to put into the base pin to switch the transistor to fully on?
Diode check... I don't know why I didn't try that. My diode check mode seems to indicate resistance, not voltage.
With red on base, putting black on emitter reads 779. Black on collector indicates 780. I'm assuming this means ohms.Black on base with red on the other pins does in fact indicate out of range, and so does any other combination that does not involve red on base.
My multimeter does have an hFE test socket, but I have no idea what I'm looking for. When I plug one of my transistors into it fluctuates erratically but stays mostly around 120. This is the current gain, right?
The transistors are pretty small. They're not the kind with a built-in heatsink. The writing says:2N4401-331
This is the Vce(voltage for collector-emitter junction) and Ie (current for collector) right? So they should be able to handle up to 30V and 800mA going through the collector-emitter junction, right?
And with a gain of 120, in order to switch it on I need to use 800/120 = ~7mA? Am I doing this correctly? So at 5V, I need a resistor of around 714Ohms? Would I fry the transistor if I were to use one of the 330ohm resistors I have lying around?? What about 1k?
This is all very confusing, but I think I'm learning and I refuse to give up.
I'll start by going backwards through your questions.Quotewith a gain of 120, in order to switch it on I need to use 800/120 = ~7mA?No. The gain is how you figure out the Ic current. Ie = Ic + IbIc = G*Ib --- G is your gain of 120 in this case.So if you put 7mA into the base of the transistor you will get 7mA*120 = 840mA through the collector/emitter junction.
Thanks so much. I'll be attempting to make an H-bridge out of them tomorrow.
If I apply just 4mA to the base then there should be no more than 4*120 = 480mA to pass through the collector-emitter junction, right? I know this would make the motors run weaker but it should keep everything from burning out right?
The second way I was thinking of is adding a resistor in series with each motor. I want no more than 500mA going through, but there's currently around 1.04A at 5V at stall. So I have a surplus of about 500mA that I need to resist. So to bring that 1A down to 500mA I need a resistor that will bring the circuit resistance to 10ohms. To verify, with the motor stalled the resistance is 5V/1A = 5Ohms. So stall resistance without a resistor is 5ohms. If I add 5 ohms in series, the circuit now has 10ohms at 5V. 5V/10 = 500mA. Am I doing this right?
But here's the part I've always struggled to understand. If I add a 5ohm resistor in series with the motor to bring its current down to a reasonable level, and the motor has 5ohms resistance itself, then the voltage drop would go from 5V across the motor to 2.5V across the resistor and 2.5V across the motor, right? So now the motor is not only at half current, but also half voltage? So it would only operate at 1/4 power? Am I understanding this correctly?
If this is true, then my only real option is to use transistors with a higher current rating, right? Then I would need to recalculate the resistance I need for the base pin. But now I know all I have to do is use the formula:(max amp rating/current gain) = current to apply to baseand(V/current to apply to base) = resistor value
As for the diode, I can just put one rectifier diode (making sure its turned the right way, of course) on the emitter pin of each transistor, right?
I do hope you enjoyed your dinner.
What's the advantage of using PNP transistors for the top? Why not simply have 4 NPN transistors, each with a resistor on its base, then connect the "forward" pin to the top left and bottom right transistors and the "backward" pin to the top right and bottom left, and use the same protective diode setup as the one in the schematic?